AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

Let the equation of parabola having horizontal axis vertex at (h,k) is,
(y−k)2=4a(x−h) .....(I)
The parabola in equation (I) passes through (−2,1) , (1, 2) and (−1, 3) so,
(k−1)2=4a(−2−h)
k2−2k+1=−8a−4ah ....(II)
Then,
(k−2)2=4a(1−h)
k2−4k+4=4a−4ah ....(III)
And,
(3−k)2=4a(−1−h)
k2−6k+9=−4a−4ah ....(IV)
From equation (II), (III) and (IV),
4a=

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